By Giuliano Sorani

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**Extra info for An introduction to real and complex manifolds.**

**Sample text**

2 log F(t) = t(X + Y ) + This proves (i). To prove (ii) put G(t) = exp(t X ) exp(tY ) exp(−t X ) exp(−tY ) t2 2 (X + 2X Y + Y 2 ) + O(t 3 ) 2 t2 · I − t(X + Y ) + (X 2 + 2X Y + Y 2 ) + O(t 3 ) 2 = I + t(X + Y ) + = I + t 2 [X, Y ] + O(t 3 ) , and (ii) follows by considering log G(t). 4 For X, Y ∈ M(n, R), lim (i) (ii) k→∞ lim k→∞ exp exp X Y exp k k k = exp(X + Y ), Y X Y X exp exp − exp − k k k k k2 = exp([X, Y ]). 3 Exercises 29 Proof. From the preceding proposition exp X Y 1 exp = exp (X + Y ) + O k k k exp X Y exp k k k = exp (X + Y ) + O 1 k2 , 1 k , and this proves (i).

Since (z) = ∞ k=0 (−1)k (z − 1)k z, k+1 we have Exp(ad X ) Exp(t ad Y ) Y = ∞ k=0 (−1)k Exp(ad X ) Exp(t ad Y ) − I k+1 k Exp(ad X ) Exp(t ad Y )Y. 5 Exercises 47 Observing that Exp(t ad Y )Y = Y, we obtain Exp(ad X ) Exp(t ad Y ) Y = ∞ k=0 (−1)k · k+1 t q1 +···+qk E(k) (ad X ) p1 (ad Y )q1 . . (ad X ) pk (ad Y )qk (ad X )m Y. q1 ! . m! The convergence of the series is uniform for t in [0, 1]. The statement is obtained by termwise integration since 1 t q1 +···+qk dt = 0 1 . 5 1 1 1 log(exp X exp Y ) = X + Y + [X, Y ] + X, [X, Y ] + Y, [Y, X ] 2 12 12 + terms of degree ≥ 4.

Then put g = exp s X and take the derivatives at s = 0, dπ ([X, Y ]) = dπ (X )dπ(Y ) − dπ(Y )dπ (X ). The adjoint representation π = Ad of G on g is a special case for which the derived representation is the adjoint representation ad of g on g. If π1 and π2 are two equivalent representations, then the derived representations dπ1 and dπ2 are also equivalent. The converse holds if G is connected.

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